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If a body is heated to $110^{\circ} \mathrm{C}$ and placed in air at $10^{\circ} \mathrm{C}$ after 1 hour its temperature is $60^{\circ} \mathrm{C}$, then the additional time required for it to cool to $30^{\circ} \mathrm{C}$ is
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Verified Answer
The correct answer is:
$\left(\frac{\log 5}{\log 2}-1\right) \mathrm{hrs}$
$\begin{aligned}
& \frac{\mathrm{d} T}{\mathrm{~d} t}=-K(T-10) \\
& \Rightarrow T-10=e^{-k t+C} \\
& \Rightarrow T=10+e^C \cdot e^{-k t}
\end{aligned}$
For $t=0$,
$\begin{aligned}
& T=110 \\
& \Rightarrow e^C=100 \text { i.e. } \\
& T=10+100 \cdot e^{-k t}
\end{aligned}$
For $t=1$,
$\begin{aligned}
& T=60 \\
& \Rightarrow 60=10+100 \cdot e^{-k \times 1} \\
& \Rightarrow-k=\log \frac{1}{2} \\
& \Rightarrow k=\log 2
\end{aligned}$
$\Rightarrow T=10+100 \cdot e^{-\left(\log _2\right) t}$
Putting $T=30$
$\begin{aligned}
& 30=10+100 \cdot e^{-\left(\log _2\right) t} \\
& \Rightarrow \log \left(\frac{1}{5}\right)=-(\log 2) t \\
& \Rightarrow t=\frac{\log 5}{\log 2}
\end{aligned}$
Additional time $=t-1=\frac{\log 5}{log2}-1$
& \frac{\mathrm{d} T}{\mathrm{~d} t}=-K(T-10) \\
& \Rightarrow T-10=e^{-k t+C} \\
& \Rightarrow T=10+e^C \cdot e^{-k t}
\end{aligned}$
For $t=0$,
$\begin{aligned}
& T=110 \\
& \Rightarrow e^C=100 \text { i.e. } \\
& T=10+100 \cdot e^{-k t}
\end{aligned}$
For $t=1$,
$\begin{aligned}
& T=60 \\
& \Rightarrow 60=10+100 \cdot e^{-k \times 1} \\
& \Rightarrow-k=\log \frac{1}{2} \\
& \Rightarrow k=\log 2
\end{aligned}$
$\Rightarrow T=10+100 \cdot e^{-\left(\log _2\right) t}$
Putting $T=30$
$\begin{aligned}
& 30=10+100 \cdot e^{-\left(\log _2\right) t} \\
& \Rightarrow \log \left(\frac{1}{5}\right)=-(\log 2) t \\
& \Rightarrow t=\frac{\log 5}{\log 2}
\end{aligned}$
Additional time $=t-1=\frac{\log 5}{log2}-1$
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